\(\frac{3}{1\cdot2}+\frac{3}{2\cdot3}+\frac{3}{3\cdot4}+...+\frac{3}{2017\cdot2018}\)
Ta có : \(=\frac{3}{1}-\frac{3}{2}+\frac{3}{2}-\frac{3}{3}+\frac{3}{3}-\frac{3}{4}+...+\frac{3}{2017}-\frac{3}{2018}\)
\(=\frac{3}{1}-\frac{3}{2018}=\frac{6051}{2018}\)
Vậy \(\frac{3}{1\cdot2}+\frac{3}{2\cdot3}+\frac{3}{3\cdot4}+...+\frac{3}{2017\cdot2018}=\frac{6051}{2018}\)
3/1.2 + 3/2.3 + 3/3.4 + ... + 3/2017.2018
= \(3.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\right)\)
= 3 . ( 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/2017 - 1/2018 )
= 3 . ( 1 - 1/2018 )
= 3 . 2017/2018
= 6051/2018
giải
3/1.2+3/2.3+3/3.4+.....+3/2017.2018
=3/1+3/2018
=6055/2018
k mình nhé
\(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{2017.2018}\)
\(=3.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\right)\)
\(=3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\right)\)
\(=3.\left(1-\frac{1}{2018}\right)\)
\(=3.\frac{2017}{2018}\)
\(=\frac{6051}{2018}\)
_Chúc bạn học tốt_
3/1.2+3/2.3+3/3.4+...+3/2017.2018
=3( 1/1.2 + 1/2.3 + 1/3.4 +...+ 1/2017.2018 )
= 3( 1 - 1/2+1/2-1/3+1/3-1/4+...+1/2017-1/2018 )
= 3( 1 - 1/2018 )
= 3 . 2017/2018
=6051/2018
Ta có :
\(\frac{3}{1.2}+\frac{3}{2.3}+...+\frac{3}{2017.2018}\)
\(=3.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2017.2018}\right)\)
\(=3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2017}-\frac{1}{2018}\right)\)
\(=3.\left(1-\frac{1}{2018}\right)\)
\(=3.\frac{2017}{2018}\)
\(=\frac{6051}{2018}\)
P/s : Đúng nha
~ Ủng hộ nhé
3/1*3 + 3/2*3 + 3/3*4 + ... + 3/2017*2018
= 3(1/1*2 + 1/2*3 + 1/3*4 + ... + 1/2017*2018)
= 3(1 - 1/2 + 1/2 + 1/3 + 1/3 - 1/4 + ... + 1/2017 - 1/2018)
= 3(1 - 1/2018)
= 3*2017/2018
= 6051/2017
Đặt A = \(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{2017.2018}\)
=> A = \(3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\right)\)
=> A = \(3.\left(1-\frac{1}{2018}\right)\)
=> A = \(3.\frac{2017}{2018}\)
=> A = \(\frac{6051}{2018}\)
